Moving #8y# to the other side of the = and change its sign
#x^2-10x+33=+8y#
To get #y# on its own divide both sides by 8
#(x^2)/8-10/8 x+33/8=y#
Write as:
#y=(x^2)/8-10/8 x+33/8#
Write as:
#y=1/8(x^2color(red)(-10)x)+33/8#
#color(green)(y_("intercept")=+33/8 larr" read directly off the equation")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(green)(x_("vertex")=(-1/2)xx(color(red)(-10)) =+5)#
The above line is part of the process of completing the square.
#color(green)("The "x^2/8"is positive so the graph is of general shape "uu)#
Substituting #x=+5# gives:
#color(green)(y_("vertex")=1/8[color(white)(./.)5^2-10(5)color(white)(.)]+33/8 =1)#
#color(green)("Vertex"->(x,y)=(5,1))#
As the graph is of general shape #uu# and #y_("vertex")# is above the x-axis there is NO X-INTERCEPT
